图中所示的圆的方程为

圆在点 $(6,16)$ 处的切线与 $y$ 轴交于点 $P$；圆在点 $(14,8)$ 处的切线与 $x$ 轴交于点 $R$。两条切线交于点 $Q$。

求四边形 $OPQR$ 面积的精确值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先由圆心和两个切点求切线，再找 $P,Q,R$。用对角线 $OQ$ 将四边形分成两个三角形，可分别用坐标轴上的边作底。

The centre is $(8,10)$. The radius gradients at $(6,16)$ and $(14,8)$ are $6/(-2)=-3$ and $-2/6=-1/3$, respectively. Therefore the tangent equations are

or $y=x/3+14$ and $y=3x-34$.

The first meets the $y$-axis at $P=(0,14)$, and the second meets the $x$-axis at $R=(34/3,0)$. Equating them gives $x/3+14=3x-34$, so $x=18$ and $y=20$. Thus $Q=(18,20)$.

Triangle $OPQ$ has base $14$ and height $18$, and triangle $OQR$ has base $34/3$ and height $20$. The exact area is

求出顶点后，也可把整个四边形看成一个梯形减去右下角的三角形。

The perpendicular radius slopes give the tangents $y=x/3+14$ and $y=3x-34$, with $P=(0,14)$, $Q=(18,20)$ and $R=(34/3,0)$. The area under the first tangent on $[0,18]$ is a trapezium with area $\frac12(14+20)(18)=306$. The triangle to the right of $RQ$ and below $Q$ has width $18-34/3=20/3$ and height $20$, giving area $200/3$. The quadrilateral area is $306-200/3=718/3$.